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UNIPLANAR MOTION OF A RIGID BODY
225
•where s is the length of the string which is unwound.   Substituting these in the energy equation we obtain
,
or = co0~ +   -y- S
= co02+270,    (a = 00)
which is the equation (3) obtained by the torque method.   Differentiating 'the last equation we have
2 co — = 2 700 at
:and
dco
= 7,
which is the equation obtained by the torque method; therefore the rest •of the problem is identical with that given by the torque method.
2. A flywheel rotates about a horizontal axis under the action of a falling body, which is suspended by means of a string wound around the axle of the flywheel. Discuss the motion, neglecting the mass of the string.
Let     I = the moment of inertia of the
rotating system.
m = the mass of the falling body. a = the radius of the axle. T = the tensile force of the string.
TOKQUE METHOD. — Taking the moments about the axis of rotation we have Q = Ta
for the resultant torque. There-f ore             Icb = Ta
is the torque equation. But considering the forces acting upon the falling body we get
mv = mg — T.
Hence
7cb = Ta
= (mg — mv) a = m (g — acib) a.
FIG. 112.
ENERGY METHOD. — Suppose the flywheel to start from rest, and let h denote the distance covered by the body during its fall. Then the energy equation gives
J Jco2 + £ mv2 = mgh.
Differentiating with respect to t,
luu + mvv = mgh. But v = au and h = v, therefore
/coco + wa2cocb = mgaco, or         Jco = m (g — acb) a.